## suss 2 years ago Xlogy-logz.ylogz-logx.zlogx-logy=1 how to prove this??

1. suss

it was supposed to x to the power logy-logz y to the power logz-logx z to the power logx-logy

2. hartnn

\(\huge x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}\) like this ?

3. suss

ah thank god..yes like that

4. hartnn

consider \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}]\) can u simplify that ?

5. hartnn

use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

6. hartnn

you should get that =0 = ln 1

7. suss

i dont understand y u multiplied the whole by log????

8. hartnn

to be able to simplify

9. hartnn

else, nothing can't be really done to that expression

10. suss

i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

11. suss

and i cudnt figure out the next step..... :(

12. hartnn

i know, i tried that also....since i could not proceed, i thought of taking log....

13. suss

if u dont mind,can u show me the simplification process...??? :)

14. hartnn

\(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}] \\\large =\log[x^{\log y-\log z}]+\log[y^{\log z-\log x}]+\log[z^{\log x-\log y}]\) got this ?

15. hartnn

now use \( \\ \log A^{B}=B\log A\)

16. hartnn

@suss

17. suss

oh sorry man i jus lost connection leeme check it out

18. suss

y did u keep the + sign ???

19. suss

@hartnn

20. hartnn

because \(\log AB=\log A+\log B \\ \)

21. suss

ah finally got it .....thx a ton man

22. hartnn

sure ? u getting that = 0 ? then write 0=log 1

23. suss

yes 100%..i hav to use formula of log ab bak ther ......i really got it man

24. hartnn

good :) welcome ^_^